$\int_0^1\frac{1}{7^{[1/x]}}dx$

Question

$$\int_0^1\frac{1}{7^{[1/x]}}dx$$ Where $[x]$ is the floor function

now as the exponent is always natural, i converted it to an infinite sum

$$\sum\limits_{k=1}^{\infty} \frac{1}{7^{[1/k]}}$$ Which is an infinite GP, whose sum is 6.

However, this is wrong, why is this wrong?

Answer 1

Note that when $x\in\left(\frac{1}{n+1},\frac{1}{n}\right],$ for some $n\in\mathbb{N},$ we have that $\lfloor\frac{1}{x}\rfloor=n.$ This observation motivates me to split up the given integral as such:

$$\begin{align*} \int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\int_\frac{1}{2}^1\frac{dx}{7}+\int_\frac{1}{3}^\frac{1}{2}\frac{dx}{7^2}+\int_\frac{1}{4}^\frac{1}{3}\frac{dx}{7^3}+\ldots\\ &=\left(\frac{1}{7}+\frac{1}{2\times7^2}+\frac{1}{3\times7^3}+\ldots\right)-\left(\frac{1}{2\times7}+\frac{1}{3\times7^2}+\frac{1}{4\times7^3}+\ldots\right)\\ &=\boxed{1+6\ln{\left(\frac{6}{7}\right)}} \end{align*}$$

I used the series expansions for $\ln(1+x)$ and $\ln(1-x)$ in the last step.

Note: Just so there is no confusion, I have more explicitly performed the calculation. From the observation, we get that $$\begin{align*} \int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\sum_{i\geq1}\int_\frac{1}{i+1}^\frac{1}{i}\frac{dx}{7^i}\\ &=\sum_{i\geq1}\left(\frac{1}{i\times7^i}-\frac{1}{(i+1)\times7^i}\right)\\ &=\sum_{i\geq1}\frac{1}{i\times7^i}-\sum_{i\geq1}\frac{1}{(i+1)\times7^i}\\ &=-\ln{\left(1-\frac{1}{7}\right)}-7\left(-\ln{\left(1-\frac{1}{7}\right)-\frac{1}{7}}\right)\\ &=\boxed{1+6\ln{\left(\frac{6}{7}\right)}} \end{align*}$$

Answer 2

More generally,

$\begin{array}\\ \int_0^1 f([1/x])dx &=-\int_1^{\infty} f([y])dy/y^2 \qquad y=1/x, x=1/y, dx=-1/y^2\\ &-=\sum_{n=1}^{\infty}\int_n^{n+1} \dfrac{f([y])dy}{y^2}\\ &=-\sum_{n=1}^{\infty}\int_n^{n+1} \dfrac{f([y])dy}{y^2}\\ &=-\sum_{n=1}^{\infty}f(n)\int_n^{n+1} \dfrac{dy}{y^2}\\ &=-\sum_{n=1}^{\infty}f(n)\dfrac1{y}\mid_n^{n+1} \\ &=-\sum_{n=1}^{\infty}f(n)(\dfrac1{n+1}-\dfrac1{n})\\ &=\sum_{n=1}^{\infty}f(n)(\dfrac1{n}-\dfrac1{n+1})\\ &=\sum_{n=1}^{\infty}\dfrac{f(n)}{n(n+1)}\\ \end{array} $

If $f(x) =\dfrac1{c^x}, c > 1, $

$\begin{array}\\ \int_0^1 f([1/x])dx &=\sum_{n=1}^{\infty}f(n)(\dfrac1{n}-\dfrac1{n+1})\\ &=\sum_{n=1}^{\infty}\dfrac1{c^n}(\dfrac1{n}-\dfrac1{n+1})\\ &=\sum_{n=1}^{\infty}\dfrac1{nc^n}-\sum_{n=1}^{\infty}\dfrac1{(n+1)c^n}\\ &=\sum_{n=1}^{\infty}\dfrac1{c^n}\dfrac1{n}-c\sum_{n=1}^{\infty}\dfrac1{(n+1)c^{n+1}}\\ &=\sum_{n=1}^{\infty}\dfrac1{nc^n}-c\sum_{n=2}^{\infty}\dfrac1{nc^{n}}\\ &=\sum_{n=1}^{\infty}\dfrac1{nc^n}-c\left(\sum_{n=1}^{\infty}\dfrac1{nc^{n}}-\dfrac1{c}\right)\\ &=(1-c)\sum_{n=1}^{\infty}\dfrac1{nc^n}+1\\ &=-(1-c)\ln(1-\frac1{c})+1\\ &=(c-1)\ln(1-\frac1{c})+1\\ \end{array} $

If $c=7$ this is $1+6\ln(\frac67) $.